Question:

<b>Second Derivative of Log Function</b><br> $y = 2\ln\!\left(\dfrac{x^2}{e^2}\right) = 2[2\ln x - 2\ln e] = 4\ln x - 4$.<br> Wait: $\log_e\!\left(\dfrac{x^2}{e^2}\right)^2 = 2\log_e\!\left(\dfrac{x^2}{e^2}\right) = 2(2\log_e x - 2) = 4\log_e x - 4$.<br> $\dfrac{dy}{dx} = \dfrac{4}{x}$, $\dfrac{d^2y}{dx^2} = -\dfrac{4}{x^2}$.<br> At $x=e$: $\dfrac{d^2y}{dx^2}=-\dfrac{4}{e^2}$. Hmm, option 2 would be $-2/e^2$... unless the function is $\ln(x^2/e^2)^1$.<br> If $y=\ln(x^2/e^2)=2\ln x-2$: $y'=2/x$, $y''=-2/x^2=-2/e^2$ at $x=e$. So option 2 = $-2/e^2$. <b>Answer: 2 = $-2/e^2$</b><br> <b>Key concept:</b> Simplify logarithm before differentiating: $\ln(x^2/e^2)=2\ln x-2$, so differentiation is straightforward.<br> <b>Trap:</b> Not simplifying first; differentiating $\ln(x^2/e^2)$ using chain rule without simplification leads to sign errors.